Respuesta :
Taking into account the definition of a system of linear equations, the solution t=14 represents a viable time that the trains are side by side.
System of linear equations
A system of linear equations is a set of two or more equations of the first degree, in which two or more unknowns are related.
Solving a system of equations consists of finding the value of each unknown so that all the equations of the system are satisfied. That is to say, the values of the unknowns must be sought, with which when replacing, they must give the solution proposed in both equations.
Calculation of time in this case
The system of equations to be solved is
[tex]\left \{ {{y=14t^{2} } \atop {y=96t+1400}} \right.[/tex]
There are several methods to solve a system of equations, it is decided to solve it using the substitution method, which consists of clearing one of the two variables in one of the equations of the system and substituting its value in the other equation.
In this case, substituting the first equation in the second you get:
14t²= 96t +1400
Solving:
14t²- 96t -1400= 0
Being this a quadratic function of the form ax² + by +c, then it can be solved by: [tex]x1,x2=\frac{-b+-\sqrt{b^{2} -4ac} }{2a}[/tex]
In this case, being a=14, b=-96 and c=-1400, the equation is solved by:
[tex]t1=\frac{-(-96)+\sqrt{(-96)^{2} -4x14x(-1400)} }{2x14}[/tex]
[tex]t1=\frac{96+\sqrt{9216 +78400} }{28}[/tex]
[tex]t1=\frac{96+\sqrt{87616} }{28}[/tex]
[tex]t1=\frac{96+296}{28}[/tex]
[tex]t1=\frac{392}{28}[/tex]
t1=14
and
[tex]t2=\frac{-(-96)-\sqrt{(-96)^{2} -4x14x(-1400)} }{2x14}[/tex]
[tex]t2=\frac{96-\sqrt{9216 +78400} }{28}[/tex]
[tex]t2=\frac{96-\sqrt{87616} }{28}[/tex]
[tex]t2=\frac{96-296}{28}[/tex]
[tex]t2=\frac{-200}{28}[/tex]
t2= -7.14
Since time cannot be a negative value, the solution t=14 represents a viable time that the trains are side by side.
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